How Does Damping Affect the Resonant Amplitude of a Driven Pendulum?

[stanli121]

Homework Statement

Given a simple pendulum with a mass on the end and a massless string. The support point for the pendulum is moved laterally with an amplitude D at the resonant frequency. The damping is from the air and is considered viscous i.e. not turbulent. The difference between the resonant frequency and the frequency in the presence of damping is negligible. What is the amplitude of the oscillation in the steady state.

Homework Equations

The Attempt at a Solution

I am trying to consider this conceptually. Because the damping is light, can I assume that the damping coefficient -> 0 and that the damping term in general cancels out? In that event, would the steady state amplitude end up being the same as the drive amplitude, D? I'm getting hung up trying to understand how the damping affects this situation because I'm told the resonant frequency is the same as the frequency in the presence of damping. Any thanks is much appreciated.

 

[JesseC]

Without damping oscillations would grow infinitely big... for a steady state to be achieved there must be some damping. The driving force is putting energy into the system, and if its not getting lost anywhere then what would happen?

 

[stanli121]

That's one of the dilemmas I encountered -- what happens to a driven, undamped oscillator. The amplitude can't grow larger than the length of the string so the other possibility seems to be a period that tends to infinity. I can't seem to grasp the fundamentals of this situation.

 

[JesseC]

To arrive at a quantitative answer the problem would require some values for the pendulum parameters and an equation relating them to the amplitude of oscillation. Such an equation does exist! I can't for the life of me remember it off by heart, but a quick search on wikipedia about damped driven harmonic oscillators might help you out.

Another problem I can see is that at large amplitude oscillations the small angle approximation will break down and the problem will become non-linear!

 

[paranormal]

I would approach this problem by first defining the variables and parameters involved. The simple pendulum system described can be represented by the equation of motion:

m(d^2x/dt^2) + c(dx/dt) + kx = F(t)

Where m is the mass of the pendulum, c is the damping coefficient, k is the spring constant, x is the displacement from equilibrium, and F(t) is the external force applied to the system.

In the case of a damped, driven oscillator, the external force is given by:

F(t) = F0sin(ωt)

Where F0 is the amplitude of the driving force and ω is the frequency of the driving force. In this problem, the resonant frequency is equal to ω, meaning that the driving force is at the system's natural frequency.

To find the amplitude of oscillation in the steady state, we can use the steady state solution to the equation of motion:

x(t) = A*sin(ωt + φ)

Where A is the amplitude of oscillation and φ is the phase shift.

Plugging this into the equation of motion and solving for A, we get:

A = F0/m(ω^2 - ω0^2)^2 + (cω)^2

Where ω0 is the natural frequency of the undamped system.

From this equation, we can see that the amplitude of oscillation in the steady state is affected by both the driving force and the damping coefficient. As the damping coefficient increases, the amplitude of oscillation decreases. However, if the damping coefficient is small, we can approximate it as 0 and the amplitude of oscillation becomes:

A ≈ F0/m(ω^2 - ω0^2)^2

In this case, the amplitude of oscillation is still affected by the driving force, but the damping term has been neglected. Therefore, the steady state amplitude would not be exactly equal to the driving amplitude, D, but it would be approximately equal if the damping is small.

In summary, the steady state amplitude of a damped, driven oscillator is affected by both the driving force and the damping coefficient. In the case of small damping, the steady state amplitude can be approximated as equal to the driving amplitude, but it is not exactly the same.